a/ \(Δ'=(-1)^2-1.(m-1)=1-m+1=2-m\)
Pt có nghiệm
\(→Δ'=2-m≥0\\↔m≤2\)
b/ Theo Vi-ét:
\(\begin{cases}x_1+x_2=2\\x_1x_2=m-1\end{cases}\)
\(x_1^3+x_2^2=(x_1+x_2)(x_1^2-x_1x_2+x_2^2)\\=2[(x_1^2+2x_1x_2+x_2^2)-3x_1x_2]\\=2[(x_1+x_2)^2-3x_1x_2]\\=2[2^2-3(m-1)]\\=8-6(m-1)\\=8-6m+6\\=14-6m\)
\(m≤2→6m≤12\\↔-6m≥-12\\→14-6m≥14-12\\↔14-6m≥2\\→\min=2\)
→ Dấu "=" xảy ra khi \(m=2\)
Vậy \(\min=2\) khi \(m=2\)